The limiting points of the co-axial system co | vedclass

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(C) The equations of the circles are $S_1: x^2+y^2+2x-2y+2=0$ and $S_2: x^2+y^2-\frac{2}{5}x-\frac{16}{5}y+\frac{13}{5}=0$. The family of circles is g

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$(-1,1), \left(\frac{1}{5}, \frac{8}{5}\right)$

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(C) The equations of the circles are $S_1: x^2+y^2+2x-2y+2=0$ and $S_2: x^2+y^2-\frac{2}{5}x-\frac{16}{5}y+\frac{13}{5}=0$. The family of circles is given by $S_1 + \lambda(S_1 - S_2) = 0$ or $S_1 + k S_2 = 0$. The limiting points are the centers of the point circles in the co-axial system. $A$ circle $x^2+y^2+2gx+2fy+c=0$ is a point circle if $g^2+f^2-c=0$. The radical axis is $S_1 - S_2 = 0$ $\Rightarrow (2 + \frac{2}{5})x + (-2 + \frac{16}{5})y + (2 - \frac{13}{5}) = 0$ $\Rightarrow \frac{12}{5}x + \frac{6}{5}y - \frac{3}{5} = 0$ $\Rightarrow 4x + 2y - 1 = 0$. Any circle in the system is $S_1 + \lambda(4x+2y-1) = 0 \Rightarrow x^2+y^2+(2+4\lambda)x + (-2+2\lambda)y + (2-\lambda) = 0$. For a point circle,$g^2+f^2-c=0 \Rightarrow (1+2\lambda)^2 + (-1+\lambda)^2 - (2-\lambda) = 0$. $1+4\lambda+4\lambda^2 + 1-2\lambda+\lambda^2 - 2+\lambda = 0 \Rightarrow 5\lambda^2+3\lambda = 0$. Thus,$\lambda = 0$ or $\lambda = -\frac{3}{5}$. For $\lambda = 0$,the center is $(-g, -f) = (-1, 1)$. For $\lambda = -\frac{3}{5}$,the center is $(-(1+2(-\frac{3}{5})), -(-1+(-\frac{3}{5}))) = (- (1-\frac{6}{5}), -(-\frac{8}{5})) = (\frac{1}{5}, \frac{8}{5})$. The limiting points are $(-1, 1)$ and $(\frac{1}{5}, \frac{8}{5})$.

The limiting points of the co-axial system containing the two circles $x^2+y^2+2x-2y+2=0$ and $25(x^2+y^2)-10x-80y+65=0$ are

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